C语言：编程求s=2/1+3/2+4/3+5/4+…+22/21的值，咋写？

Question

如题所述

Answer 1

代码文本：

#include "stdio.h"

int main(int argc,char *argv[]){

int i;

double s;

for(s=0.0,i=1;i<22;s+=1.0/i++);

printf("2/1+3/2+4/3+...+22/21 = %f\n",s+21);

return 0; 

}

Answer 2

注意int到float的强制转换

#include "stdafx.h"

#include <iostream>

using namespace std;

int main()

{

float s = 0, t;

for (int i = 1; i <= 21; i++)

{

t = (float)(i + 1) / i;

s += t;

}

cout << s << endl;

system("pause");

    return 0;

}

Answer 3

#include <stdio.h>

int main()

{

float i, S = 0;

for (i = 1; i <=21; i++)

S = S + (i+1)/i;

printf("%.2f\n", S);

return 0;

}